[tex]\displaystyle\\
\text{Folosim formula:}\\\\
a^0+a^1+a^2+a^3+\cdots +a^n=\frac{a^{n+1} -1}{a-1},~~\text{unde }a \in N\\\\
\text{In enuntul dat lipseste }~3^0=1\\
\text{Din acest motiv, din formula de mai sus vom scadea }1=3^0\\\\
3+3^2+3^3+\cdots 3^x=120\\\\
\frac{3^{x+1}-1}{3-1}-1=120\\\\
\frac{3^{x+1}-1}{2}=120+1\\\\
\frac{3^{x+1}-1}{2}=121\\\\
3^{x+1}-1=121\cdot2\\
3^{x+1}-1=242\\
3^{x+1}=242+1\\
3^{x+1}=243\\
\text{Observam ca }243=3^5\\
3^{x+1}=3^5\\
x+1=5\\
x=5-1\\
\boxed{\bf x=4}[/tex]
